The value of ' $a$ ' for which the function

$f(x)=\left\{\begin{array}{cl}\frac{1-\cos 4 x}{x^2}, & x<0 \\ \frac{a}{\sqrt{x}}, & x=0 \text { is continuous at } x=0, \text { is } \\ \frac{\sqrt{16+\sqrt{x}}-4}{\sqrt{16+}} & \end{array}\right.$

If $\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots \ldots \infty$ and $\mathop {\lim }\limits_{x \to 0} \frac{\log (1+x)^{1+x}}{x^2}-\frac{1}{x}=k$, then $12 k=$

If $f(x)=\left\{\begin{array}{ll}k, & \text { for } x=1 \\ \frac{(9 x-1)(\sqrt{x}-1)}{3 x^2+2 x-5}, & \text { for } x \neq 1\end{array}\right.$ is continuous on $[0, \infty)$, then $k=$

In each of the choices given below, a function and an interval are given. The correct choice having a function and the associated interval for which the Lagrange's mean value theorem is not valid is